Formula for Cochran's Critical Values

  • Thread starter Thread starter Dave Curtis
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Dave Curtis

Hi,

I'm trying to do some data analysis using Cochran's test for outlying
variances.
I have 4 replicate numbers from each of 20 laboratories. I calculate the
variance of each set of data.
I can work out the Cochran's test value by dividing the maximum variance by
the sum of all the variances.
Then I need to compare this with the Cochran critical values, which are
available from tables, but these have gaps, so I'd like to be able to
calculate them.
Does anyone know of a formula to calculate these?

Dave
 
Dave,

You could try to use linear interpolation:

For a value in A2, with your table in D2:EXXX, and values in D are what should match A2:

=TREND(OFFSET($E$2,MATCH(A2,$D$2:$D$XXX)-1,0,2,1),OFFSET($D$2,MATCH(A2,$D$2:$D$XXX)-1,0,2,1),A2)

HTH,
Bernie
MS Excel MVP
 
it doesn't look like there's a simple formula for small samples, although it
approaches a Chi squared for larger ones (cf.
http://www.watpon.com/table/cochran.pdf).
Somewhat more accurate than a linear approximation would be to use cubic
interpolation around the neighbouring points eg for k=50 and v=1:

=TREND(B15:B18,A15:A18^{1,2,3},A25^{1,2,3})

gives 0.2599 as opposed to 0.2461 for the linear case. This formula can be
generalised by adapting Bernie's formula above (using offset(...-2,0,4,1) and
^{1,2,3}).
 
Thanks for the info.
I was hoping to be able to replicate Cochran's values with a formula, but
I've been unable to ascertain how they were derived.
Lori, your idea of a cubic interpolation seems a good one. I've only done
linear interpolations before. However, using your formula, I get a value of
0.2461, instead of the 0.2599 you obtain. Which bracketing points are best
for a cubic interpolation?
I'm not a statistician, so I'm groping in the dark a little here.

Thanks

Dave
 
Dave, i think you're right - it was a typo. It's best to use the neighbouring
points for this ie between the interval BC use the points ABCD, at the
endpoints you can use the two before or after.
 
Dave

I have devlopped a spreadsheet that includes a macro that will claculate the Cochran value for any combination of sets and dgrees of freedom.

I used Cochran's original paper (1941) to test it and also tested it against published tables.

If you want a copy of the spreadshhet leet me know.

Lou



DaveCurti wrote:

Formula for Cochran's Critical Values
22-Jan-09

Hi

I'm trying to do some data analysis using Cochran's test for outlying
variances
I have 4 replicate numbers from each of 20 laboratories. I calculate the
variance of each set of data
I can work out the Cochran's test value by dividing the maximum variance by
the sum of all the variances
Then I need to compare this with the Cochran critical values, which are
available from tables, but these have gaps, so I'd like to be able to
calculate them.
Does anyone know of a formula to calculate these

Dave

Previous Posts In This Thread:

Formula for Cochran's Critical Values
Hi

I'm trying to do some data analysis using Cochran's test for outlying
variances
I have 4 replicate numbers from each of 20 laboratories. I calculate the
variance of each set of data
I can work out the Cochran's test value by dividing the maximum variance by
the sum of all the variances
Then I need to compare this with the Cochran critical values, which are
available from tables, but these have gaps, so I'd like to be able to
calculate them.
Does anyone know of a formula to calculate these

Dave

it doesn't look like there's a simple formula for small samples, although it
it doesn't look like there's a simple formula for small samples, although it
approaches a Chi squared for larger ones (cf.
http://www.watpon.com/table/cochran.pdf).
Somewhat more accurate than a linear approximation would be to use cubic
interpolation around the neighbouring points eg for k=50 and v=1

=TREND(B15:B18,A15:A18^{1,2,3},A25^{1,2,3}

gives 0.2599 as opposed to 0.2461 for the linear case. This formula can be
generalised by adapting Bernie's formula above (using offset(...-2,0,4,1) and
^{1,2,3})

:

Thanks for the info.
Thanks for the info
I was hoping to be able to replicate Cochran's values with a formula, but
I've been unable to ascertain how they were derived
Lori, your idea of a cubic interpolation seems a good one. I've only done
linear interpolations before. However, using your formula, I get a value of
0.2461, instead of the 0.2599 you obtain. Which bracketing points are best
for a cubic interpolation
I'm not a statistician, so I'm groping in the dark a little here

Thank

Dav

:

Dave, i think you're right - it was a typo.
Dave, i think you're right - it was a typo. It's best to use the neighbouring
points for this ie between the interval BC use the points ABCD, at the
endpoints you can use the two before or after

:


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Dave

I have devlopped a spreadsheet that includes a macro that will claculate the Cochran value for any combination of sets and dgrees of freedom.

I used Cochran's original paper (1941) to test it and also tested it against published tables.

If you want a copy of the spreadshhet leet me know.

Lou
Click to expand...

You could post it onto a free hosting site like 'mediafire' or the
like, then post the link here.

OR, you could post it as a template on the Microsoft template site.
That would only be if it is macro free, or if you put all the macros into
a worksheet as text, allowing the user to apply them into the VBeditor
manually to get the workbook to function.
 
Maybe you find this article helpful:

R.U.E. 't Lam, "Scrutiny of variance results for outliers: Cochrans test optimized", Analytica Chimica Acta 659 (2010) 68-84.
Reprints available through (e-mail address removed)

In this article I develop an equation (Eq. 28) that will calculate accurate critical values Cc for the traditional Cochran's C test using Excel. The equation requires critical F values Fc as input parameter. Fc can be obtained from Excel function FINV: FINV[significance level, degrees of freedom 1, degrees of freedom 2].

Cc(alpha1,n,L) = 1 / {1 + (L-1) / FINV[alpha1/L, (n-1), (L-1)(n-1)]}

Where
alpha1 = one-sided significance level
L = total number of data series
n = total number of replicates in a single data series

Equation 28 works for any significance level 0 <= alpha1 <= 1, any number of data series L >= 2, and any number of replicates per data series n >= 2.

 
Last edited:
Blog "Variance Outlier Test"

I have started a blog with
- Short introduction to the Variance Outlier Test (G test)
- Additional explalantion
- My updated manuscript
- More extensives tables with critical values

http://www.rtlam.blogspot.com/
 
Hi,

I'm trying to do some data analysis using Cochran's test for outlying
variances.
I have 4 replicate numbers from each of 20 laboratories. I calculate the
variance of each set of data.
I can work out the Cochran's test value by dividing the maximum variance by
the sum of all the variances.
Then I need to compare this with the Cochran critical values, which are
available from tables, but these have gaps, so I'd like to be able to
calculate them.
Does anyone know of a formula to calculate these?

Dave
Click to expand...

A formula to calculate critical values for Cochran’s C test can be found in:

R.U.E. ’t Lam, "Scrutiny of variance results for outliers: Cochran’s test optimized", Analytica Chimica Acta 659 (2010) 68–84. Equation 28 calculates the exact critical values for the traditional Cochran's C test. Theformula works for any number of data sets, any number of replicates per data set, and at any confidence level. The equation requires critical F as input parameter. Critical F is obtained from Excel function FINV.

To make the contents of the article readily available, and to give further directions on how to perform this variance outlier test, I maintain a blog:
http://rtlam.blogspot.com/

Regards,

Ruben 't Lam
 
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